Problem: The polynomial $p(x)=x^3-6x^2+32$ has a known factor of $(x-4)$. Rewrite $p(x)$ as a product of linear factors. $p(x)=$
Solution: We know $(x-4)$ is a factor of $p(x)$. This means that $p(x)=(x-4)\cdot q(x)$ for some polynomial $q(x)$. We can find $q(x)$ using polynomial division, and then we can factor $q(x)$. This way, we will be able to rewrite $p(x)$ as a product of linear factors. Dividing $p(x)$ by $(x-4)$ Notice that $p(x)$ is missing a $1^{\text{st}}$ degree term. Let's add it as $0x$. $\begin{array}{r} x^2-2x-\phantom{1}8 \\ x-4|\overline{x^3-6x^2+0x+32} \\ \llap{--(}\underline{x^3-4x^2\phantom{+0x+32}\rlap)} \\ -2x^2+0x+32 \\ \llap{--(}\underline{-2x^2+8x\phantom{+32}\rlap)} \\ -8x+32 \\ \llap{--(}\underline{-8x+32\rlap)} \\ 0 \end{array}$ We find that $q(x)=x^2-2x-8$. Factoring $q(x)$ We can factor $x^2{-2}x{-8}$ as $(x+m)(x+n)$ where $m+n={-2}$ and $m\cdot n={-8}$. Such numbers are $2$ and $-4$, so the factored expression is $(x+2)(x-4)$. Putting it all together $\begin{aligned} p(x)&=x^3-6x^2+32 \\\\ &=(x-4)(x^2-2x-8) \\\\ &=(x-4)(x+2)(x-4) \end{aligned}$ Notice that we can also write $p(x)$ as $(x-4)^2(x+2)$. This is still a product of linear factors!